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Proof step 6 from proof bib-on-bits (previous step)

¬(# int2bits(i) ≤ 8 ∧ int2bits(i) ≠ []), ~128 ≤ i, i ≤ 127 ⊦ i = bits2int(lastn(8, fillsgn(8, int2bits(i))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
8 < # int2bits(i), i ≤ 127, ~128 ≤ i ⊦ i = bits2int(lastn(8, int2bits(i)))

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ n < # bits → lastn(n, fillsgn(n, bits)) = lastn(n, bits);% ⊦ int2bits(j) ≠ [];% ⊦ ¬ m ≤ n ↔ n < m;